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problem:
Find the limit. (If you need to use -http://www.webassign.net/images/infinity.gif or http://www.webassign.net/images/infinity.gif, enter -INFINITY or INFINITY.)
http://www.webassign.net/cgi-bin/sym...%204%20x%29%29
I would explain what ive done in more detail, but i know im missing something due to x approaching neg. infinity, but i did multiply the above funct. by its conjugate to get (x-1)/(x-square root of x^2+4x). How would i go about this? Thank you. -Evan
Let $\displaystyle x = -t$ and calculate $\displaystyle \lim_{t \rightarrow + \infty} \left( \sqrt{ t^2 - 4t} - t\right)$.Quote:
Originally Posted by Evan.Kimia
Ok here is what i have...
sqroot(t^2-4t)-t * sqroot(t^2-4t)+t
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1 * sqroot(t^2-4t)+t
t^2-4t-t^2
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sqroot(t^2-4t)+t
consolidate like terms then divide by t....
-4
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* sqroot(t^2-4t)+1
how can i properly solve for the denominator with the sqroot involved? I have something on paper that gets me sqroot(t-4)+1 which is most likely wrong. Thanks.
When you divided the denominator by t, you failed to divide the square root term by t. The correct expression is $\displaystyle \frac{-4}{\sqrt{1 - \frac{4}{t}} + 1}$. Now take the required limit.Quote:
Originally Posted by Evan.Kimia
I see. How would i properly divide t by the denominator? I know this is easy, but my algebra is off. I know dividing by t in the sqroot wont really work unless t itself is within a square root, so t= sqroot(t^2), am i right? If thats so, then i can see sqroot(t^2-4t)
turning into (1-4/t), but wouldnt the 1 turn into 1/sqroot(x^2) which would equal 0? I know im wrong, just thinking out loud.
Also, you explained this better than my math teacher today, thanks for everything.
You divide each term in the denominator by t.Quote:
Originally Posted by Evan.Kimia
Dividing the square root term by t works in the way you have described. And when you divide the other term by t (and that other term is t) then you get t/t = 1. So you end up with a denominator as I posted.
